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Countable or Uncountable…That is the question! REVIEW • Countable – Empty set, finite set or countably infinite • Countably Infinite – The set is a non-empty, non-finite set, and there exists a bijection between N and the set. • Uncountable – Not countable HOMEWORK Solutions (1) Yes The function f(n) = 2n is the desired bijection. (2) Yes The function f(n) = desired bijection. -n/2 if n is even (n-1)/2 if n is odd is the Are the Rational Numbers Countable? • What do we know about rational numbers? ••••••- VOTING Are the Rational Numbers countable? (A) YES (B) NO (C) UNSURE What about the interval (0,1) • What do we know about this interval? ••••••- VOTING Reals in the interval (0,1) countable? (A) YES (B) NO (C) UNSURE Let’s prove some things to attack these questions! • If A & B are disjoint countably infinite sets then AυB is countable. Proof • Since A is countable there exists a bijection f : N A such that f (i) = ai • Since B is countable there exists a bijection g : N B such that g (i) = bi • Construct a function h (i) that orders the elements of A and B in the following way: a1, b1, a2, b2, a3, b3, . . . Our function h (i) • h (i) = • Why is h a bijection? RECORD these observations on your worksheet Let’s prove some things to attack these questions! • If A & B are disjoint countably infinite sets then AυB is countable. • If A is a countably infinite set and B is a subset of A then B is countable. If A is a countably infinite set and B is a subset of A then B is countable. Case I: If B is the empty set or a finite set then B is countable. Case II: B is an infinite set Since A is countable we can write the elements of A in the order a1, a2, a3, . . . If B is a subset of A then an infinite number of elements in the above sequence are elements of B. Thus the elements of B form a subsequence (c1, c2, c3,. . .) of the sequence a1, a2, a3, . . ., thus we may order the elements of B as b1, b2, b3, . . . where bk = ck and the function f (i) = bi is a bijection between N and B Let’s prove some things to attack these questions! • If A & B are disjoint countably infinite sets then AυB is countable. • If A is a countably infinite set and B is a subset of A then B is countable. • How does (N x N) relate to Q+ υ {0} in size? . . . . . . . . . . . . . . . 0,2 1,2 2,2 3,2 . . . 0,1 1,1 2,1 3,1 . . . 0,0 1,0 2,0 3,0 . . . . . . . . . . . . . . . . . f(1) = (0,0) . f(2) = (1,0) 0,2 1,2 2,2 3,2 . . . f(3) = (0,1) f(4) = (0,2) f(5) = (1,1) 0,1 1,1 2,1 3,1 . . . f(6) = (2,0) f(7) = (3,0) 0,0 1,0 2,0 3,0 . . . f(8) = (2,1) f(9) = (1,2) . . . Prove It! • Now that we know that NxN is countable we can show that Q is countable. • Use the facts we have deduced to show that Q is countable Proof that Q is countable • We know that Q+ υ {0} can be thought of as a subset of NxN • Similarly Q- can be thought of as a subset of NxN • Q+ υ {0} and Q- are countable because they are subsets of a countable set. • We have shown that the union of two countable sets is also countable so (Q+ υ {0}) υ Q- = Q is countable Hey! Q is countable! • Does this change your mind about the real numbers in the interval (0,1) being countable/uncountable? (A) YES (B) NO (C) UNSURE Cantor’s Diagonalization Argument • Cantor proved that the interval of real numbers (0,1) is… UNCOUNTABLE!!! • We start by noting that each real number in the interval (0,1) has a unique decimal representation of the form 0.d1d2d3d4… (where each di is a number from 0-9). And where decimals with period 1 cannot repeat with the number 9. Proof (by contradiction) • Assume that f is a bijection from N (0,1). Then we may say: f(1) f(2) f(3) f(4) = = = = A = 0.a1a2a3a4… B = 0.b1b2b3b4… (Where A,B,C,D are distinct real numbers in (0,1)) C = 0.c1c2c3c4… D = 0.d1d2d3d4… . . . Choose a digit from 0 to 8 and we will call this a’1 such that a’1 ≠a1 Similarly choose a digit from 0 to 8 for b’1 such that b’2≠ b2 Similarly choose c’3≠ c3 and d’4≠ d4 . . . Clearly f does not map a natural number to the real number 0.a’1b’2c’3d’4… So f is not a bijection. Contradiction! • This proof show us that the interval (0,1) is actually a larger size of infinity than the natural numbers. HENCE (0,1) is a larger size of infinity than Q as well! Homework • Question #1 – Today we identified two kinds of infinity: the size of the natural numbers and the size of the interval (0,1). – Show that there is a bijection between the interval (0,1) and the set of real numbers. – What does the existence of this bijection imply about these two sets? • Question #2 – Can you find a different size of infinity? That is a set that cannot be put into a bijection with N or the interval (0,1). – To help you with this problem research the findings of Paul Cohen.